Finding the order of $\langle a,b | a^{8}=b^{2}=1, ab=ba^{3}\rangle.$

Question

Im new at abstract algebra stuff and im wondering whats the technique to prove this kind of stuff.

Question:

Let $G=\langle a,b | a^{8}=b^{2}=1, ab=ba^{3}\rangle$, prove that $|G|=16 $ and find all subgroups or order $8$.

I can calculate all the elements of $G$, then I showed by brute force that $G$ has only $16$ elements but I cant give a real proof of that.

Any help will be really appreciated, thanks for the help. :D

Answer 1

It is easy to see that $G$ can have at most $16$ elements, because we can list all elements in the form $b^sa^r$ with $0\le r\le 7$ and $0\le s\le 1$ by using $ab=ba^3$. On the other hand, such a group is given by $SD_2$, the semidihedral group of degree $2$. For further details see the classifcation of all groups of order $16$ here.